Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x+4y &= 2 \\ 6x+4y &= -1\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $6x = -4y-1$ Divide both sides by $6$ to isolate $x$ $x = {-\dfrac{2}{3}y - \dfrac{1}{6}}$ Substitute this expression for $x$ in the first equation. $2({-\dfrac{2}{3}y - \dfrac{1}{6}}) + 4y = 2$ $-\dfrac{4}{3}y - \dfrac{1}{3} + 4y = 2$ Simplify by combining terms, then solve for $y$ $\dfrac{8}{3}y - \dfrac{1}{3} = 2$ $\dfrac{8}{3}y = \dfrac{7}{3}$ $y = \dfrac{7}{8}$ Substitute $\dfrac{7}{8}$ for $y$ in the top equation. $2x+4( \dfrac{7}{8}) = 2$ $2x+\dfrac{7}{2} = 2$ $2x = -\dfrac{3}{2}$ $x = -\dfrac{3}{4}$ The solution is $\enspace x = -\dfrac{3}{4}, \enspace y = \dfrac{7}{8}$.